The voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of 10⁶ V m⁻¹. The plate area is 10⁻⁴ m². What is the dielectric constant if the capacitance is 15 pF? (Takeε₀=8.86×10⁻¹² C² N⁻¹ m⁻²)

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(2) 8.5. The capacitor is used right up to its rating, so the maximum field appears across the plates: d=V/(Eₘₐₓ)=(500)/(10⁶)=5×10⁻⁴ m. Without the dielectric, C₀=(ε₀A)/d=(8.86×10⁻¹²×10⁻⁴)/(5×10⁻⁴)=1.772×10⁻¹² F=1.772 pF. C=KC₀⇒ K=(15)/(1.772)=8.47. So K≈8.5.

The voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of 10⁶ V m⁻¹. The plate area is 10⁻⁴ m². What is the dielectric constant if the capacitance is 15 pF? (Takeε₀=8.86×10⁻¹² C² N⁻¹ m⁻²)

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